Find an explicit formula for the geometric sequence $120\,,\,60\,,\,30\,,\,15,...$. Note: the first term should be $\textit{a(1)}$. $a(n)=$
Solution: In a geometric sequence, the ratio between successive terms is constant. This means that we can move from any term to the next one by multiplying by a constant value. Let's calculate this ratio over the first few terms: $\dfrac{15}{30}=\dfrac{30}{60}=\dfrac{60}{120}={\dfrac{1}{2}}$ We see that the constant ratio between successive terms is ${\dfrac{1}{2}}$. In other words, we can find any term by starting with the first term and multiplying by ${\dfrac{1}{2}}$ repeatedly until we get to the desired term. Let's look at the first few terms expressed as products: $n$ $1$ $2$ $3$ $4$ $g(n)$ ${120}\cdot\!\left({\dfrac{1}{2}}\right)^{\,0}$ ${120}\cdot\!\left({\dfrac{1}{2}}\right)^{\,1}$ ${120}\cdot\!\left({\dfrac{1}{2}}\right)^{\,2}$ ${120}\cdot\!\left({\dfrac{1}{2}}\right)^{\,3}$ We can see that every term is the product of the first term, ${120}$, and a power of the constant ratio, ${\dfrac{1}{2}}$. Note that this power is always one less than the term number $n$. This is because the first term is the product of itself and plainly $1$, which is like taking the constant ratio to the zeroth power. Thus, we arrive at the following explicit formula (Note that ${120}$ is the first term and ${\dfrac{1}{2}}$ is the constant ratio): $a(n)={120}\cdot\left({\dfrac{1}{2}}\right)^{{\,n-1}}$ Note that this solution strategy results in this formula; however, an equally correct solution can be written in other equivalent forms as well.